Linear Equations: Area of Triangle Formed by Three Lines

Calculate the area of the triangle formed by the pairwise intersections of three lines using geometric decomposition.

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Slope-Intercept Form & Function Notation

When three distinct lines intersect pairwise at three distinct points, they form a triangle. To determine its area, we find the three intersection vertices and apply geometric decomposition with a reference right triangle.

Triangle Area Decomposition Visualizer

Lines:

L_1: y = 0, \quad L_2: y = -2x + 8, \quad L_3: y = 2x

Triangle vertices:

A(0,0), B(4,0), C(2,4)

Additive Area Calculation:

S_{\Delta ABC} = S_{ADC} + S_{BDC} = \frac{2 \cdot 4}{2} + \frac{2 \cdot 4}{2} = 4 + 4 = 8

1. Calculate the Three Vertices

Solve pairwise linear systems to determine the three intersection coordinates:

A(x_A, y_A)

B(x_B, y_B)

, and

C(x_C, y_C)

2. Construct a Reference Right-Angled Triangle

Draw a horizontal or vertical auxiliary line through one vertex to construct a right-angled reference triangle alongside two additional triangles.

Additive Method

Given vertices $A(0,0)$ , $B(4,0)$ , $C(2,4)$ :

Auxiliary Line: An auxiliary vertical line through $C(2,4)$ intersects base $AB$ at $D(2,0)$ .
Sub-regions: This divides the triangle into two adjacent right triangles:
- $S_1 = \triangle ADC$ : base $AD = 2$ , height $CD = 4 \implies S_1 = \frac{2 \cdot 4}{2} = 4$
- $S_2 = \triangle BDC$ : base $BD = 2$ , height $CD = 4 \implies S_2 = \frac{2 \cdot 4}{2} = 4$
Calculation: $S_{\Delta ABC} = S_1 + S_2 = S_{ADC} + S_{BDC} = 4 + 4 = 8$

Subtractive Method 1: Reference Triangle Subtraction

Given vertices $A(0,0)$ , $B(4,4)$ , $C(3,1)$ :

Reference Right Triangle: Construct auxiliary point $D(4,0)$ on the x-axis to form reference right-angled triangle $\triangle ABD$ with base $AD = 4$ , height $BD = 4$ , and area $S_{\text{ref}} = \frac{4 \cdot 4}{2} = 8$ .
Triangles to Subtract:
- $S_1 = \triangle BCD$ : base $BD = 4$ , height $= 1 \implies S_1 = \frac{4 \cdot 1}{2} = 2$
- $S_2 = \triangle ACD$ : base $AD = 4$ , height $= 1 \implies S_2 = \frac{4 \cdot 1}{2} = 2$
Calculation: $S_{\Delta ABC} = S_{\text{ref}} - S_1 - S_2 = S_{\Delta ABD} - S_{\Delta BCD} - S_{\Delta ACD} = 8 - 2 - 2 = 4$

Subtractive Method 2: Bounding Box Subtraction

Given vertices $A(1,1)$ , $B(5,2)$ , $C(2,5)$ :

Bounding Box: Enclose the triangle inside a bounding box from $(1,1)$ to $(5,5)$ with width $4$ , height $4$ , and area $S_{\text{box}} = 4 \cdot 4 = 16$ .
Exterior Triangles to Subtract:
- $S_1$ : base $= 4$ , height $= 1 \implies S_1 = \frac{4 \cdot 1}{2} = 2$
- $S_2$ : base $= 3$ , height $= 3 \implies S_2 = \frac{3 \cdot 3}{2} = 4.5$
- $S_3$ : base $= 1$ , height $= 4 \implies S_3 = \frac{1 \cdot 4}{2} = 2$
Calculation: $S_{\Delta ABC} = S_{\text{box}} - S_1 - S_2 - S_3 = 16 - 2 - 4.5 - 2 = 7.5$

Learning Topics

Frequently Asked Questions

What is the slope-intercept form of a linear equation?

The slope-intercept form is $y = ax + b$, where **$a$ is the slope** and **$b$ is the y-intercept**.

How do we express $y$ as a function of $x$?

When we rewrite a linear equation to isolate $y$ in the form $y = ax + b$, the value of $y$ is determined entirely by $x$. This allows us to use function notation and write $y = f(x)$.